The lifespans of bears in a particular zoo are normally distributed. The average bear lives $32.8$ years; the standard deviation is $5.8$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a bear living between $44.4$ and $50.2$ years.
Solution: $32.8$ $27$ $38.6$ $21.2$ $44.4$ $15.4$ $50.2$ $99.7\%$ $95\%$ $2.35\%$ $2.35\%$ We know the lifespans are normally distributed with an average lifespan of $32.8$ years. We know the standard deviation is $5.8$ years, so one standard deviation below the mean is $27$ years and one standard deviation above the mean is $38.6$ years. Two standard deviations below the mean is $21.2$ years and two standard deviations above the mean is $44.4$ years. Three standard deviations below the mean is $15.4$ years and three standard deviations above the mean is $50.2$ years. We are interested in the probability of a bear living between $44.4$ and $50.2$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the bears will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $95\%$ of the bears will have lifespans within 2 standard deviations of the mean. That leaves $99.7\% - 95\% = 4.7\%$ of bears between 2 and 3 standard deviations of the mean, or $2.35\%$ on either side of the distribution. The probability of a particular bear living between $44.4$ and $50.2$ years is $\color{orange}{2.35\%}$.